A compound contains only C, H, N, and O . Combustion of a 1.48-g sample in excess O2 yields 2.60

A compound contains only C, H, N, and O . Combustion of a...

A compound contains only $\mathrm{C}, \mathrm{H}, \mathrm{N},$ and $\mathrm{O} .$ Combustion of a $1.48-\mathrm{g}$ sample in excess $\mathrm{O}_{2}$ yields $2.60 \mathrm{~g} \mathrm{CO}_{2}$ and $0.799 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .$ A separate experiment shows that a $2.43-\mathrm{g}$ sample contains $0.340 \mathrm{~g} \mathrm{~N}$. What is the empirical formula of the compound?

Key Concepts

Empirical Formula Determination

This concept involves finding the simplest whole?number ratio of the elements in a compound. It is a crucial step in chemical analysis, where experimental data (often masses or percentages of elements) are converted, typically via moles, into a ratio that reflects the identity of the compound in its simplest form.

Combustion Analysis

Combustion analysis is a key technique for determining the amounts of carbon, hydrogen, and sometimes oxygen in organic compounds. In this process, a compound is burned in excess oxygen to produce CO? and H?O, which are then quantified to determine the original amounts of carbon and hydrogen present in the compound.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It is used to convert masses of products (like CO? and H?O) to moles of elements by using their molar masses, allowing for the determination of the proportions of each element in the original compound.

Elemental Analysis and Mass Balance

Elemental analysis is the process of determining the elemental composition of a substance. When a compound contains additional elements like nitrogen, its content is established through separate experiments. The oxygen content can then be deduced by applying a mass balance approach, subtracting the masses of the determined elements from the total mass of the compound.

Transcript

00:01 In this question, we've been given some information relating to a compound which contains carbon, hydrogen, nitrogen and oxygen.

00:09 So looking at the information that we've been given, we've been told that 1 .48 grams of the compound, it yields 2 .6 grams of co2 and 0 .799 grams of h2 when bent in xsa.

00:26 So we've been asked to look into its empirical formula.

00:30 So looking at the information that we've been given here, the first thing that we'd want to do is to convert the number of moles, rather to convert the masses of carbon dioxide and h2o into number of moles.

00:42 Recall that number of moles is equal to the mass divided by the molar mass.

00:46 So number of moles of co2, these are going to be equal to 2 .6 divided by 44, which is going to give us 0 .0591, moles of co2 and the number of moles of h2o these are going to be 0 .7 .99 divided by 1 8 which is going to give us 0 .444 0 .04 for 3 moles of h2o.

01:10 Now that we have these number of moles, one more of carbon dioxide it gives us 1 more of carbon and 2 moles of oxygen and 1 more of h2o it gives us 2 moles of hydrogen and 1 more of oxygen.

01:26 So looking at these ratios right here, we are saying the number of most of carbon, they are equal to the number of moles of co2, which is equal to 0 .0591 modes.

01:37 Therefore, the mass of carbon, this is going to be equal to 0 .091 multiplied by 12, which gives us a mass of carbon that is equal to 0 .71 claims of carbon.

01:50 If we do the same thing for hydrogen, the number of moles of hydrogen, they are equal to 2 ,000.

01:54 Multiplied by the number of moles of h2o, which is equal to 2 multiplied by 0 .0443.

02:03 Therefore, the mass of hydrogen, this is 2 multiplied by 0 .0443 by 1, which is the atomic weight of hydrogen.

02:13 And this gives us a mass of hydrogen that is going to be equal to 0 .0 895 grams of hydrogen.

02:24 Now that we have this information, we can then determine the mass of nitrogen...

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